The volume of H2 obtained at S.T.P. when Mg obtained by passing a current of 0.5 amp through molten MgCl2 for 32.2 minutes is treated with excess dilute HCl is approximately [Eq. mass of Mg = 12)

Question

The volume of H2 obtained at S.T.P. when Mg obtained by passing a current of 0.5 amp through molten MgCl2 for 32.2 minutes is treated with excess dilute HCl is approximately [Eq. mass of Mg = 12) (A) 56 cm3 (B) 28 cm3 (C) 5.6 cm3 (D) 112 cm3

Tardigrade Admin · Accepted Answer

Mass of Mg deposited = Zlt = (12/96500) × 0.5 × 32.2 × 60 = 0.12012 g Now, <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/chemistry/26518f7b17184e531438c823072270c8-.png" />

Q. The volume of $H_{2}$ obtained at S.T.P. when $M g$ obtained by passing a current of $0.5$ amp through molten $M g C l_{2}$ for $32.2$ minutes is treated with excess dilute $H Cl$ is approximately [Eq. mass of $M g = 12$ )

56 $c m^{3}$

28 $c m^{3}$

5.6 $c m^{3}$

112 $c m^{3}$

Mass of Mg deposited = Zlt
$= \frac{12}{96500} \times 0.5 \times 32.2 \times 60$
= $0.12012 g$
Now,

Q. The volume of H2​ obtained at S.T.P. when Mg obtained by passing a current of 0.5 amp through molten MgCl2​ for 32.2 minutes is treated with excess dilute HCl is approximately [Eq. mass of Mg=12)

56 cm3

28 cm3

5.6 cm3

112 cm3